Lecture 14: Basic Number Theory

Nitin Saxena ?

IIT Kanpur

1 Euler’s totient function φ

The case when n is not a prime is slightly more complicated. We can still do modular arithmetic with division if we only consider numbers coprime to n. For n ≥ 2, let us deﬁne the set,

∗ Zn := {k | 0 ≤ k < n, gcd(k, n) = 1} . ∗ The cardinality of this set is known as Euler’s totient function φ(n), i.e., φ(n) = |Zn|. Also, deﬁne φ(1) = 1.

Exercise 1. What are φ(5), φ(10), φ(19) ?

Exercise 2. Show that φ(n) = 1 iﬀ n ∈ [2]. Show that φ(n) = n − 1 iﬀ n is prime.

Clearly, for a prime p, φ(p) = p − 1. What about a prime power n = pk? There are pk−1 numbers less than n which are NOT coprime to n (Why?). This implies φ(pk) = pk − pk−1. How about a general number n? We can actually show that φ(n) is an almost multiplicative function. In the context of number theory, it means,

Theorem 1 (Multiplicative). If m and n are coprime to each other, then φ(m · n) = φ(m) · φ(n) .

∗ ∗ ∗ ∗ ∗ Proof. Deﬁne S := Zm × Zn = {(a, b): a ∈ Zm, b ∈ Zn}. We will show a bijection between Zmn and ∗ ∗ ∗ ∗ ∗ S = Zm × Zn. Then, the theorem follows from the observation that φ(mn) = |Zmn| = |S| = |Zm||Zn| = φ(m)φ(n). ∗ The bijection ψ : S → Zmn is given by the map ψ :(a, b) 7→ bm + an mod mn. We need to prove that ψ is a bijection. That amounts to proving these three things.

∗ ∗ ∗ – The mapping is valid, i.e., if a ∈ Zm and b ∈ Zn then bm + an ∈ Zmn. This follows from the fact that bm is coprime to n implies bm + an is coprime to n. Similarly bm + an is coprime to m. So bm + an is coprime to mn (and we use its residue representative in [mn − 1]). – Mapping ψ is injective (one to one). Why? If bm + an = b0m + a0n mod mn implies (b − b0)m + (a − a0)n = 0 mod mn. The latter implies, using coprimality of m, n, that n|(b − b0) and m|(a − a0). Thus, (a, b) = (a0, b0) in S. – Mapping ψ is surjective (onto). Why? ∗ −1 ∗ Consider t ∈ Zmn. Compute k := tm mod n. (Note: k ∈ Zn.) Since t = km mod n we can write 0 0 0 ∗ t = km + `n. If need be, reduce ` to ` mod m. This achieves both t = km + ` n mod mn and ` ∈ Zm. These three properties of ψ ﬁnish the proof.

Exercise 3. Find numbers m, n such that φ(mn) 6= φ(m)φ(n).

? Edited from Rajat Mittal’s notes. Fundamental theorem of arithmetic implies that we can express any number as a product of prime powers. By using Thm 1, we can calculate φ(mn), when φ(m) and φ(n) are given to us (m and n are coprime).

k1 k2 k` Theorem 2. If n = p1 p2 ··· p` is a natural number. Then, 1 1 1 φ(n) = n 1 − 1 − ··· 1 − . p1 p2 p` Exercise 4. Prove the above theorem using the argument above.

2 Inclusion-Exclusion vs. M¨obiusInversion

There is another way to look at Thm. 2. We are interested in finding out the number of elements between 0 k1 k2 k` and n−1 which do not share a factor with n = p1 p2 ··· p` . Let us consider all the elements {0, 1, ··· , n−1}. Define Ai to be the set of elements which are divisible by pi. For any I ⊆ [`], define AI to be the set of elements which are divisible by all pi where i ∈ I. You can see that we are interested in the event when none of the pi’s, where i ∈ [`], divide an element. This is a straightforward application of inclusion-exclusion,

X |I| φ(n) = (−1) · |AI | . I⊆[`] n Notice that the number of elements which are divisible by p1p2 ··· pj is just . This gives us, p1p2···pj n |AI | = Q . i∈I pi So, X n φ(n) = (−1)|I| . (1) Πi∈I pi I⊆[`] Exercise 5. Prove that the above expression is the same as the one in Thm. 2.

In Eqn. 1, the sum is taken over all square-free (i.e. of the form p1p2 ··· pi with distinct primes) divisors of n. Deﬁne a function, µ(k),   1, if k = 1 µ(k) := 0, if a2 | k for some a ≥ 2  (−1)r, if k is square-free with r primes. This function µ(k) is called the M¨obiusfunction. Then Eqn. 1 can be rewritten as, X n φ(n) = µ(d) · . d d|n P

Exercise 6. For an integer n ≥ 2 show that, d|n µ(d) = 0 .

] ` [ ∈ i ] ` [ ∈ i n | d i i

i i

)=0. = )) p ( µ + (1 = )) p ( µ + · · · + ) p ( µ + ) p ( µ + (1 = ) d ( µ that deduce

Q Q P 2 i k

.Ti a eue to used be can This ). ab ( µ = ) b ( µ · ) a ( µ , b a, coprime for i.e. function, multiplicative a is ) k ( µ

Möbiusfunction is really useful in number theory, and combinatorics. One of the main reasons is the “inversion property” (for special functions f). Theorem 3 (Möbiusinversion). Let f and g be functions defined on natural numbers. Then, X X n f(n) = g(d) implies g(n) = µ(d)f . d d|n d|n

2 Proof. Let us look at RHS of the expression for g(n):

X n X X µ(d)f = µ(d) g(c) d d|n d|n c|n/d   X X = g(c) µ(d) c|n d|n/c = g(n)µ(1) = g(n) . P The third equality follows from the fact that d|n µ(d) is 0 for n ≥ 2 (is 1 for n = 1). The second equality is sum-swapping.

Exercise 7. Prove the second equality by considering the pairs (c, d) s.t. d | n and c | n/d.

Functions f and g are called M¨obiustransforms of each other. Eg. n and φ(n) are M¨obiustransforms of each other!

Exercise 8. Finite fields are routinely used in computer science. Read up on how to use Möbiusinversion to count the number of irreducible polynomials, of degree d, over a finite field.

References

1. N. L. Biggs. Discrete Mathematics. Oxford University Press, 2003. 2. P. J. Cameron. Combinatorics: Topics, Techniques and Algorithms. Cambridge University Press, 1994. 3. K. H. Rosen. Discrete Mathematics and Its Applications. McGraw-Hill, 1999.